Советы от нейросети в категории period

Solving for Time and Frequency of a Wave

2023-12-23 12:02:35

To solve this problem, we need to first find the solutions to the given quadratic equation 4x2-20x+25=0. This can be done using the quadratic formula, which states that the solutions to the equation ax2+bx+c=0 are given by x=(-b±√(b^2-4ac))/2a. In this case, a=4, b=-20, and c=25, so the solutions are x=5/2 and x=5. However, since we are looking for a time and not just a value for x, we will only use the positive solution x=5. Now, we know that the time it takes for a wave to complete one full cycle is equal to the period, T, which is given by T=1/f, where f is the frequency. In our case, since the wave is completing 900 cycles, we can set T=900/f. Since we also know that the time it takes for the wave to complete one cycle (the period) is equal to 5 seconds (from our solution for x), we can set 5=900/f and solve for f. This gives us a frequency of 180 Hz. Therefore, the time it will take for the wave to complete 900 cycles is 5 seconds, and the frequency of the wave is 180 Hz.

Решение задачи о скорости и периоде обращения искусственного спутника на круговой орбите

2023-12-12 09:52:36

Скорость искусственного спутника для круговой орбиты на высоте 400 км над земной поверхностью должна быть приблизительно 7,6 км/с. Это скорость, которая позволяет спутнику оставаться на одной орбите без падения на поверхность планеты. Чтобы вычислить период его обращения, можно использовать закон Кеплера: T = 2π * √(a^3/GM), где T - период обращения, a - большая полуось орбиты, G - гравитационная постоянная, M - масса Земли.

В данном случае, a = 700 км (400 км над истинным поверхностным радиусом Земли в 6371 км), G = 6,674 * 10^-11 м^3/кг*с^2, M = 5,972 * 10^24 кг. Подставляя значения в формулу, получаем период обращения спутника вокруг Земли равным примерно 92,5 минуты (1 час 32 минуты 30 секунд).

Calculating the frequency of oscillatory motion

2023-12-11 21:13:56

The frequency of the body's oscillatory motion can be easily calculated using the formula f = 1/T, where T is the period of oscillation. In this case, f = 1/20 с = 0.05 Hz .

Calculating Centripetal Acceleration for Two Particles on Circular Motion

2023-11-09 21:32:01

In the case of equal speeds, the centripetal accelerations of the two point masses will be different due to the difference in their respective radii.

The first step to solving this problem is to determine the speed, v, of the particles. Using the formula for circular motion v = 2πR/ T, where R is the radius and T is the period of rotation, we can express the speed in terms of R and T.

When comparing the two centripetal accelerations, we can plug in the value of v into the formula for centripetal acceleration, a = v^2/R. In this case, the accelerations will be proportional to the ratio of the radii, as v is the same for both particles. Since it is given that R₁ = 2R₂, we can conclude that the centripetal acceleration of the first particle will be twice that of the second particle.

For the second case, where the periods of rotation are equal, we can use the same formula for circular motion, v = 2πR/T, to express the speeds in terms of the radii and periods. When looking at the formula for centripetal acceleration, a = v^2/R, we can see that the accelerations will be directly proportional to the square of the radii, as v and T are both the same for both particles. Therefore, the first particle with radius R₁ will have a centripetal acceleration four times that of the second particle with radius R₂.